Dana Russo

2022-09-07

$$\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2}$$

I think to do :

$$\frac{(\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2})}{\sqrt{2}}$$

but i dont get anything. Or to divied by $\sqrt{3}$ :

$$\frac{(\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2})}{\sqrt{3}}$$

I think to do :

$$\frac{(\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2})}{\sqrt{2}}$$

but i dont get anything. Or to divied by $\sqrt{3}$ :

$$\frac{(\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2})}{\sqrt{3}}$$

barquegese2

Beginner2022-09-08Added 11 answers

Hint:

$$\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}\frac{\pi}{3}\mathrm{sin}x-\mathrm{cos}\frac{\pi}{3}\mathrm{cos}x=\frac{\sqrt{2}}{2}$$

$$\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}\frac{\pi}{3}\mathrm{sin}x-\mathrm{cos}\frac{\pi}{3}\mathrm{cos}x=\frac{\sqrt{2}}{2}$$

tonan6e

Beginner2022-09-09Added 2 answers

One of the R-Formulas, a set of formulas for combining such trigonometric expressions, says that

$$a\mathrm{sin}x-b\mathrm{cos}x=R\mathrm{sin}(x-\alpha )$$

where

$$R=\sqrt{{a}^{2}+{b}^{2}},\alpha ={\mathrm{tan}}^{-1}\frac{b}{a}$$

However, I doubt this solution expresses any room for creativity for this question specifically, as noted by first answer

$$a\mathrm{sin}x-b\mathrm{cos}x=R\mathrm{sin}(x-\alpha )$$

where

$$R=\sqrt{{a}^{2}+{b}^{2}},\alpha ={\mathrm{tan}}^{-1}\frac{b}{a}$$

However, I doubt this solution expresses any room for creativity for this question specifically, as noted by first answer

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